Find k such that $3 ^{- x / 2}$ = $\mathrm { e } ^ { \mathrm { kx } }$ for all x.
A) - $\frac { 1 } { 2 }$ ln 3
B) ln 3
C) $\frac { 1 } { 2 }$ ln 3
D) ln - $\frac { 3 } { 2 }$

Question

Quizplus · Accepted Answer

To find the value of $k$ that makes the equation $3^{-x/2} = e^{kx}$ true for all $x$, we can equate the exponents of the bases $3$ and $e$ by converting the base $3$ to base $e$ using the natural logarithm. The conversion is done as follows: $3^{-x/2} = (e^{\ln 3})^{-x/2} = e^{-x/2 \cdot \ln 3}$. For the equation to hold for all $x$, the exponents on both sides must be equal, which gives us $kx = -x/2 \cdot \ln 3$. Dividing both sides by $x$ (assuming $x \neq 0$) gives $k = -\frac{1}{2} \ln 3$.

Find K Such That 3−x/23 ^{- x / 2}3−x/2 = ekx\mathrm { e } ^ { \mathrm { kx } }ekx

Find K Such That $3 ^{- x / 2}$ = $\mathrm { e } ^ { \mathrm { kx } }$